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tekilochka [14]
3 years ago
15

1. How much force does it take to move a 500kg box with an acceleration of 2 m/s/s?

Physics
1 answer:
Mashcka [7]3 years ago
5 0

f = ma

f = 500 × 2 = 1,000.

Answer: 1,000 N (Newtons)

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An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o
Vesnalui [34]

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

cos(\alpha) = 80/320 = 0.25

\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0 relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h

7 0
3 years ago
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Two planets have masses 2 x 10^23 kg and 5 x 10^22 kg, and the distance between them is 3 x 10^16 m. Calculate the gravitational
xxTIMURxx [149]

Explanation:

given,

mass of one planet (m1)=2*10^23 kg

mass of another planet (m2)=5*10^22kg

distance between them(d)=3*10^16m

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gravitational force between them(F)=?

we know,

F=Gm1m2/d^2

or, F=6.67*10^-11*2*10^23*5*10^22/(3*10^16)^2

or, F=6.67*2*5*10^-11+23+22/3*3*10^32

or, F=66.7*10^34/9*10^32

or, F=7.41*10^34-32

•°• F=7.41*10^2

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3 0
3 years ago
video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It
natka813 [3]

Answer:

\omega_f = 0.4\ rad/s

Explanation:

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width of door dimension  = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

I_{total} =I_{door} + I_{bullet}

I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2

a) from conservation of angular momentum  

L_i = L_f  

mv\dfrac{W}{2} = I_{total}\omega_f  

mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f

\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}

\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}

\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}

\omega_f = 0.4\ rad/s

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3 years ago
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Answer:D

Explanation:

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Answer:

B ) Ascend using my buddy alternative air source / make an emergency Ascent

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